# Subring of a finitely generated Noetherian ring need not be Noetherian? [duplicate]

A common example showing that a subring of a Noetherian ring is not necessarily Noetherian is to take a polynomial ring over a field $k$ in infinitely many indeterminates, $k[x_1,x_2,\dots]$. The quotient field is then obviously Noetherian, but the subring $k[x_1,x_2,\dots]$ is not since there is an infinite ascending chain of ideals which never stabilizes.

Is there an instance of a finitely generated Noetherian ring over some ground ring $R$, that has an intermediate ring which is not finitely generated over $R$, and hence not Noetherian either?

## marked as duplicate by Lord Shark the Unknown abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 27 '18 at 4:24

• This is answered by Mariano's answer here. That question did not have a Noetherian condition, but it's clearly satisfied in the example. – Dylan Moreland Feb 5 '12 at 22:23
• Oh yeah, Hilbert's basis theorem shows the Noetherian condition. Should I delete this somehow? – Emilia Feb 5 '12 at 22:32
• I actually don't know what the interface for asking questions is like, but if you want to then that seems fine: no one has written an answer, or anything. – Dylan Moreland Feb 5 '12 at 22:46

To get this question off the unanswered list, I copy Mariano's example from here.

The polynomial ring $k[x,y]$ is Noetherian (by Hilbert's basis theorem), but the subring generated by $\{xy^i:i\geq0\}$ is not finitely generated over $k$.

• Nice, simple example. – anomaly Dec 15 '15 at 20:09